Question

Drivers with an average of 20/40 vision travel at 55 mph in the curb lane of a freeway, where exit ramps are designed for 25 mph. What should be the minimum distance of signs with 6-in. letters placed ahead of the exit?
The following information may be useful:
Perception-reaction time 2.5 seconds;
Deceleration rate - 5 ft/sec2:
Road level-1% (downgrade);
Drivers with 20/20 vision can read signs at 60 ft per inch of letter height. .

Answer 1

Answer:

The sign board must be placed 573 ft ahead of the exit.

Explanation:

Distance needed for reducing the speed from 55 mph to 25 mph is given as

$d=\frac{v_f^2-v_i^2}{30 \times (\frac{a}{g}-G)}$

Here

• $v_f$ is the velocity at the end which is 25 mph
• $v_i$ is the velocity at the start which is 55 mph
• a is the rate of deceleration which is -5 ft/s^2
• G is the Road grade which is 1% or 0.01
• g is the gravitational acceleration whose value is 32.2 ft/s^2

$d=\frac{v_f^2-v_i^2}{2 \times (\frac{a}{g}-G)}\\d=\frac{25^2-55^2}{30 \times (\frac{-5}{32.2}-0.01)}\\d=551 ft$

Now the perception time is 2.5 second, 20/20 vision  person can read 6 inch letters from 60 x 6 ft.

For 20/40 vision person can read 6 inch letters from 30 x 6 ft=180 ft.

SSD is $d=\frac{55 \times 5280 \times 2.5}{60 \times 60}\\d=202 ft$

So the minimum distance is given as

$Minimum distance=551+202-180 ft\\Minimum distance=573 ft$

So the sign board must be placed 573 ft ahead of the exit.